Let $\pi: E \rightarrow M$ be a fiber bundle over the manifold M and denote by $\Gamma(E)$ the space of smooth sections of $E$.

For compact $M$ it is well known (Hamilton 1982, Part II Corollary 1.3.9), that $\Gamma(E)$ (if not empty) is a (tame) Fréchet manifold with respect to the topology of uniform convergence of all derivates on compacta. E.g. the topology is given by seminorms (shown here for vector bundles):
$$p_{i, K} (\phi) = \sum_{j=1}^i \text{sup}_{x \in K} |\phi^{(j)}(x)|.$$
Where the section $\phi$ is identified with its local representative $\phi: U \subset R^n \rightarrow R^m$ and the compact sets $K$ form a exhaustion of $U$. As for a paracompact manifold there exists a countable atlas, this procedure results in countable many seminorms. Thus $\Gamma(E)$ is a Fréchet manifold. (For the general case of a fiber bundle one has to invoke the tubular neighborhood theorem.)

I`m now interested in the case of non-compact base manifold $M$. To be honest, I do not see why the above construction fails then.

Supportive to this view, in section 2.2. of [1] the authors construct along the above lines a topology for non-compact $M$. But on the other hand in [2] the gauge group $\text{Gau}(P)$ (which is the group of sections of the associated bundle $P \times_G G$ to the principal bundle $P \rightarrow M$) is described only as a strict inductive limit of countable many Fréchet spaces and only for compact $M$ one has the simpler Fréchet structure on $\text{Gau}(P)$.

I thought the gauge group was the space of sections of $M \times G$. $P \times_G G$ is naturally isomorphic to $P$ unless I've misunderstood.
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Paul ReynoldsDec 16 '12 at 21:24

Hmm maybe the fact is (I didn't read the articles, so I'm just guessing) that for a non compact base manifold, the Fréchet structures you obtain are not tamely equivalent...
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SamueleDec 16 '12 at 22:47

@PR $G$ is acting on itself by conjugation, not translation in $P\times_G G$. So eg $P\times_G G$ always has sections, and $P$ never does if $P$ is non-trivial.
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PaulDec 17 '12 at 1:32

3 Answers
3

Your definition depends on the choice of the exhaustion and on the choice of the metric on $E$. To get a meaningful theory you have to add many more assumptions (like: a Riemannian metric of bounded geometry on $M$ where the open sets are geodesic balls ...).
For example, if you want to let the diffeomorphism group of $M$ act smoothly on the locally convex space of functions you are defining.
Just keep in mind how many different function spaces on $\mathbb R^n$ are useful.

Both references model on spaces of test functions (Choice 1 in the answer of Andrew Stacey below). There are other choices, but they are increasingly complicated. See for example the following paper which discusses the group of diffeomorphisms on $\mathcal R^n$ which fall rapidly towards the identity, or fall like $H^\infty$ (intersection of all Sobolev spaces).

Peter W. Michor and David Mumford: A zoo of diffeomorphism groups on $\mathbb ℝ^n$. arXiv:1211.5704. (pdf)

Thank you for pointing out the dependence on the construction. Do you know any independence-results if I fix a (pseudo-)Riemannian metric (even Lorentzian manifold if necessary) on the base space and restrict to affine or vector bundles (to include the additive structure Andrew Stacey pointed out)?
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Tobias DiezJan 2 '13 at 10:52

Another enquiry regarding the dependence on the exhaustion and the metric in the non-compact case, as I do not see it. I suspect you are referring to the seminorms $p_{K, k}(f) = sup_{x \in K}(|\nabla^i f (m) |)$. Let $(K_n)$ and $(L_i)$ be two different exhaustions by compact sets. By a standard argument each $L_i$ can be absorbed by one $K_n$, $\L_i \subseteq K_n$. Than trivially $p_{L_i} \leq p_{K_n}$ and the topology is independent on the choice of exhaustion. Furthermore, the metric always gets evaluated over compact sets and so each choice of metric should give equivalent topologies.
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Tobias DiezMar 26 '13 at 18:26

For non-compact base you need to take a limit over compact subspaces. So when considering two sections, say $f$ and $g$, then to say that they are "close" is to say that there is a neighbourhood of $f$ containing $g$. This neighbourhood says "There is a compact subset, say $K$, of $M$ and an order, say $n$, such that the derivatives of $f$ and $g$ are close on $K$ up to order $n$."

This says absolutely nothing about what happens outside $K$. And this is a big problem because if you want to put the structure of a manifold on $\Gamma(E)$ then for $f$ and $g$ sufficiently close you need to be able to say what $f + g$ is. On $K$ then there's no problem because you make everything sufficiently small that you can use the manifold structure of $E$ to add $f$ and $g$ fibrewise. But to extend that to $x$ outside $K$ you need to be able to add $f(x)$ and $g(x)$ where these can take any values in the fibre at $x$, which equates to a fibrewise global addition structure on $E$ and in a general fibre bundle you don't have that.

Now, you could go for a different source for your local additive structure but you'd run in to the same sort of problem.

You have two options if you want to work with $\Gamma(E)$ as a smooth object:

Change your topology. You can split $\Gamma(E)$ into pieces where $f$ and $g$ are in the same piece if they agree off some compact subset. This then can be made into a manifold but it has uncountably many components.

Change your category. The space $\Gamma(E)$ is a perfectly well behaved generalised smooth space and can be treated very nicely in one of the many categories of such. It just isn't a manifold.

Thanks for this clear and vivid explanation, I think I got the point! Do you know any results, if I have such a additive structure on $E$ available (e.g. vector or affine bundles)?
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Tobias DiezJan 2 '13 at 10:58

If your fibres are vector/affine bundles then the space of sections will be a vector space/affine space and then you can treat it as such. Then (I think) you get a locally convex topology from your construction (if M is sigma-compact then the topology is Frechet, I think) so you have a locally convex topological vector space and you can work with that as a smooth space. So it is a manifold, but for slightly the wrong reason!
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Loop SpaceJan 2 '13 at 12:08

where they study how to introduce a Banach structure on sets of maps between a possibly non-compact topological space as domain and a smooth manifold as target. Of course the regularity discussed here is less than $C^\infty$, giving you a Banach structure instead of Frechet. This has obvious advantages and disadvantages; but regardless of the differences is possibly worth looking at, given that the non-compactness issues of the domain are dealt with successfully. Needless to say, once you have the desired structure on the set of maps, restricting to the particular case of sections of a bundle is straight-forward.